For the following compare [1]}. Call \(\operatorname{supp}(\psi )=\bigcap _{\psi ^{\prime }\in [\psi ]}\overline{\lbrace \psi ^{\prime }\ne 0\rbrace }\)  the support of \(\psi \) . There is \(\psi ^{\prime }\in [\psi ]\)  satisfying \(\operatorname{supp}(\psi )=\overline{\lbrace \psi ^{\prime }\ne 0\rbrace }\) .Generally, let \(\Delta \subset \mathbb {R}^d\)  be measurable and set \(C:=\bigcap \lbrace \overline{\Delta \setminus N}: N \textrm { null set}\rbrace \) ,
\(C^{\prime }:=\lbrace x\in \mathbb {R}^d: \lambda (\Delta \cap U)>0\,\forall \, U \textrm { open, } x\in U\rbrace \) , and \(C^{\prime \prime }:=\mathbb {R}^d\setminus \bigcup \lbrace U \textrm { open}: \lambda (U\cap \Delta )=0\rbrace \)  with \(\lambda \)  the Lebesgue measure. Then \(C=C^{\prime }=C^{\prime \prime }\) . Moreover, \(N:=\Delta \setminus C\)  is a null set and \(C=\overline{\Delta \setminus N}\) . — Indeed, one easily checks \(C^{\prime }=C^{\prime \prime }\) . If \(x\in C^{\prime }\)  then \(\lambda (\Delta \cap U)>0\)  for all open \(U\)  with \(x\in U\) . This implies that \((\Delta \setminus N) \cap U\ne \emptyset \)  for every null set \(N\) . Hence \(x\in \overline{\Delta \setminus N}\)  for all \(N\) . Therefore \(x\in C\) . Thus \(C^{\prime }\subset C\) . For the proof of the reverse inclusion consider any countable base \(\mathcal {B}\)  of \(\mathbb {R}^d\)  and put \(V:=\bigcup \lbrace U\in \mathcal {B}: \lambda (U\cap \Delta )=0\rbrace \) . Obviously \(\Delta \cap V\)  is a null set. Let \(N\supset \Delta \cap V\)  be any null set and let \(x\in \overline{\Delta \setminus N}\) . Then for every open \(U\)  with \(x\in U\)  there is \(x^{\prime }\in (\Delta \setminus V)\cap U\) . One infers \(\lambda (\Delta \cap U)>0\) . Hence \(x\in C^{\prime }\) . Thus \(C\subset C^{\prime }\) . Finally, note \(V=\mathbb {R}^d\setminus C\) . So \(N:=\Delta \setminus C\)  is a null set. Therefore \(C\subset \overline{\Delta \setminus N}=\overline{C\cap N}\subset C\) , whence the last part of the assertion. — The question remains open whether \(C\setminus \Delta \)  is a null set.
